1. Finite Groups S13.4

Spring 2013 Problem 4: Let G be a group with an odd number of elements that has a normal subgroup N with 17 elements. Show that N lies in the center of G.

1.1. Ideas S13.4.

• If N is a normal subgroup then conjugation gives a map from G to Aut(H).

• This map is trivial iff N is central.

• The only group of order 17 is C_{17}.

• There are 16 automorphisms of C_{17}.

• The only map from a group of odd order to one of order 16 is trivial.

1.2. Write up S13.4. Claim:

(1) ρ : G → Aut(H) given by ρ(g)(h) = ghg^{−1} is a group homomorphism.

(2) The order of the image of a homomorphism of finite groups divides both the order of the domain and that of the range.

(3) (|G|, |Aut(N)|) = 1.

(4) N lies in the center of G iff the image of ρ is trivial. By (1), (2) and (3) the image of ρ is trivial and so by (4) N is central. Justifications:

(1) This is an easy computation.

(2) The image is a subgroup of the range and a quotient group of the domain so these divisibilities are a standard lemma.

(3) By the classification of prime order groups N is cyclic and hence any automorphism is determined by the image of a generator and since the order is prime there is an automorphism taking the generator to any element except the identity. Thus there are 17 − 1 automorphisms. Since |G| is odd (|G|, 16) = 1.

(4) This is a definition of the center: z is in the center iff for every g ∈ G one has gzg^{−1} = z.

2. Finite Groups F14.2

Fall 2014 Problem 2. Let G be a finite group and N ⊆ G be a subgroup of index p, where p is the smallest prime dividing |G|. Prove N is a normal subgroup of G.

2.1. Ideas F14.2.

• Primes suggest Sylow subgroups for which there are normality criteria.

• Consider the action of G on some set such as:

• (a) conjugates of a subgroup

• (b) cosets of a subgroup

• (a)(i) conjugates are more related to normality.

• (a)(ii) For contradiction even if N is not normal the kernel K of this action ρ is and maybe relates to N...

• (a)(iii) The stabilizer of the trivial conjugate N contains N and hence if (for contradiction) N is not normal then the stabilizer T is exactly N. This has made use of the prime index hypothesis.

• (a)(iv) The image of ρ is a quotient of G and a subgroup of Σ_{p} Oops; how do I know that the number of conjugates of N is p? (cosets is obvious).

• (a)(iv)(α) This folllows from (a)(iii) since the index of the stabilizer of an element of a transitive action is the number of elements

• (a)(iv)(β) The only number dividing p! with p as the smallest prime factor is p so the order of the image of ρ which is the index of K is also p and K ⊆ T = N so K = N is normal.

2.2. Write up F14.2. Consider ρ : G → S with S the symmetric group on {N^{h} = hNh^{−1}} (the conjugates of N in G) given by ρ(g)(N^{h}) = N^{gh}. Write K = ρ^{−1}(e_{S}) for the kernel and T for the stabilizer of N = N^{e_{G}}.

Claim:

(1) ρ is a group homomorphism.

(2) K is normal in G.

(3) K ⊆ T.

(4) N ⊆ T.

(5) (p!, |G|) = p.

(6) [K : G]|p.

(7) T = G iff N is a normal subgroup of G. If K = G then T = G and N is normal. Otherwise K = T = N and again N is normal.

Justifications:

(1) This is an easy computation.

(2) The kernel of any homomorphism is normal.

(3) This is immediate.

(4) This is immediate.

(5) p is the smallest prime dividing |G|.

(6) The index divides both |G| and p! and hence p by (5).

(7) This is a definition of normal: for every n ∈ N and g ∈ G one has gng^{−1} ∈ N so ρ(g)(N^{e}) = N^{e}.

3. Finite Groups F16.2

Fall 2016 Problem 2: Let G be a finite group that acts transitively on a set X of cardinality ≥ 2. Show that there exists an element g ∈ G which acts on X without any fixed points. Is the same true if G is infinite?

3.1. Ideas F16.2.

• If x ∈ X there is a subgroup G^{x} fixing x.

• There is also a subset G^{xy} of group elements taking x to y.

• The transitivity gives a bijection between G^{x} = G^{xx} and each other G^{xy} so that G = |X||G^{x}|. This is not meaningful if G is infinite.

• The goal is to show that ∪G^{x} is not all of G.

• The map from G to S_{X} might as well be injective since otherwise the quotient of any counterexample by the kernel of this map yields another counterexample.

• One example is A_{4} in S_{4} which does not have a cycle of length 4.

• Another example is C_{4} in S_{4} for which every G^{x} is trivial.

• Maybe Sylow groups are useful.

• For the infinite case the permutations of ℕ which fix all sufficiently large numbers is a counterexample.

• Another infinite counterexample is SO(3) acting on the two sphere.

• Going back to the counting: the union of the G^{x} has fewer than |G^{x}||X| elements since there are overlaps.

4. Finite Groups S16.4

Spring 2016 Problem 4: Prove that every finite group G of order > 2 has a nontrivial automorphism.

4.1. Ideas S16.4.

• If G is not Abelian there is a nontrivial inner automorphism (by conjugation).

• If G is finite and Abelian it is isomorphic to a product of cyclic groups.

• If H has a nontrivial automorphism so does H × K.

• Every finite cyclic group has a nontrivial automorphism except for C_{2}.

• C_{2} × C_{2} has a nontrivial automorphism.